If X is path connected how may i show that the reduced Suspension $\Sigma $ X is then simply connected. I cannot seem to picture this construction
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2Do you know the seifert can kampen theorem? – mland Sep 13 '12 at 07:01
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Whilst I thought it was true immediately by the Freudenthal suspension theorem, there are a serious of posts by Tom Goodwillie http://www.lehigh.edu/~dmd1/tg26, http://www.lehigh.edu/~dmd1/tg27, http://www.lehigh.edu/~dmd1/tg28 that suggests it might not (always be true). Although I can't see any flaw in the argument provided below... – Juan S Sep 13 '12 at 11:38
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@JuanS: I haven't read the posts in detail, but in the first one the counterexample he produces involved the Hawaiian earring which does not satisfy the hypotheses of Freudenthal's suspension theorem. – Bruno Stonek Mar 10 '14 at 12:31
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An overkilling answer could be the following: The Freudenthal suspension theorem tells us that, if $X$ is $n$-connected, then the natural morphism
$$ \pi_k(X) \longrightarrow \pi_{k+1}(\Sigma X) $$
is an isomorphism for $k\leq 2n$. Particularly, for $n=0$, we have an isomorphism $\pi_1(\Sigma X) = 1$.
But, if you want to "picture" the situation, take a look at this suspension drawing, and use the Seifert-van Kampen theorem, as mland points you. Particularly, look at Wikipedia's computation of $\pi_1(S^2)$.
Agustí Roig
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Well, this doesn't say why $\pi_0(\Sigma X)=0$ :) (which can be seen just from the definition of $\Sigma X$). – Bruno Stonek Jan 14 '15 at 16:22
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Here's a brief outline:
Note that $\Sigma X = X \times I / ( X \times \partial I \cup \{x_0\} \times I )$, where $x_0$ is assumed to be the base point of $X$ and $\partial I = \{0,1\}$.
Taking any loop $f: [0,1] \to \Sigma X$, one can define the following homotopy: $$ H(x, t) = [(f(x), 1-t)] $$ where for any $\alpha \in X \times I$, $[\alpha]$ denotes its equivalence class in $\Sigma X$.
Use the universal property of quotient spaces to show that this map is continuous, and check that it is also a homotopy. This will show that every loop is homotopic to the constant loop, and thus, $\pi_1(X) = 0$.
As for connectedness of $\Sigma X$, just note that $X$ is connected, so $X \times I$ is connected, and quotients of connected spaces are connected, therefore $\Sigma X$ is connected.
